Respuesta :

[tex]\bf 19sin(x)+9=sin(x)\implies 19sin(x)-sin(x)=-9 \\\\\\ 18sin(x)=-9\implies sin(x)=\cfrac{-9}{18}\implies sin(x)=-\cfrac{1}{2} \\\\\\ sin^{-1}[sin(x)]=sin^{-1}\left( -\cfrac{1}{2} \right)\implies \measuredangle x=sin^{-1}\left( -\cfrac{1}{2} \right) \\\\\\ \measuredangle x = \begin{cases} \frac{7\pi }{6}\\\\ \frac{11\pi }{6} \end{cases}[/tex]
ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE